Issue 021 - Astronomy - Orbital scaling
How fast does Betelgeuse's companion orbit?
Astronomers using ESO's Very Large Telescope have obtained the clearest image yet of a likely companion star orbiting Betelgeuse. Reuters reported a separation of about 8.8 Earth-Sun distances and a system mass near 20 solar masses. Use Earth as the reference orbit.
The problem
Estimate the companion's orbital period, its approximate orbital speed, and how many complete orbits astronomers could have observed during the roughly century in which a companion has been suspected.
Use Earth's orbit as your reference case: radius 1 Earth-Sun distance, central-system mass 1 solar mass, period 1 year, and speed roughly 30 km/s.
Given the orbital period, is it surprising that astronomers needed about a century to confirm the companion directly?
Because Fermi problems target an order of magnitude, I normally use no more than two significant digits and write most calculations in scientific notation; the Fermi reference explains both conventions.
Before checking sources
Matt's first pass
I do not have great models for orbital velocity, but I assumed the main force involved would be gravity between the bodies. The formula I remembered was something like:
gravitational force ~= G x (mass 1 x mass 2) / distance^2I figured the orbital velocity would need to increase or decrease according to the difference in that gravitational force.
For the Earth-Sun system, I used an Earth mass of about 6 x 10^23 kg and a solar mass of about 2 x 10^30 kg, so the top of the equation is about 1.2 x 10^54. For the Betelgeuse system, I treated the combined mass as 20 times the Earth-Sun system and the distance as about 9 AU.
Earth reference force scale ~= 1
Betelgeuse force scale ~= 20 / 9^2
~= 20 / 81
~= 1/4Looking back, I could have just used 1 for the Earth system and 20/81 for Betelgeuse, without bringing in my memory pegs for Earth and Sun masses.
If it is fair to say one-quarter gravitational force means one-quarter the rate of revolutions, then 4 years in the Earth system corresponds to 1 year in the Betelgeuse system. So in a century, it would have orbited about 25 times.
Then I took the rough speed given for Earth, 30 km/s, multiplied by 3 x 10^7 seconds to get the circumference of Earth's orbit, divided by 2 pi to get the radius, multiplied that radius by 9, multiplied by 2 pi again to get the Betelgeuse path, and divided by 12 x 10^7 seconds, four years, to get an orbital speed of about 7 km/s.
Calibration Score
Matt's Calibration Score: 30 / 100
Higher is better: earn points for accurate pegs, sound models, correct math, and a result close to the sourced answer. The image shows percent full of it: 100 minus the Calibration Score.
Pegs: 10/30. The mass and distance anchors were usable, but the orbital-physics peg was missing.
Model: 0/30. Gravitational-force scaling alone is not the right model; Kepler scaling is the needed tool.
Math: 10/10. The arithmetic was mostly clean once the chosen model was set.
Result: 10/30. The period was still in the broad neighborhood, but the speed estimate suffered.
Grounding facts
Saturn orbits the Sun at about 9.5 AU and takes about 29 years, but Betelgeuse's companion at roughly 8.8 AU takes only about 6 years because the system is about 20 times more massive than the Sun.
That comparison is the lesson: orbital distance pushes periods longer, but central mass pulls periods shorter. In this case, the high mass wins enough to make the orbit fast on human timescales.
After checking sources
Check and recalibrate
The right shortcut is Kepler's third law. If period is measured in Earth years, distance in AU, and mass in solar masses, then:
P^2 ~= a^3 / MHere, use a ~= 8.8 AU and M ~= 20 solar masses:
a^3 ~= 8.8^3
~= about 9^3
~= 7.3 x 10^2
P^2 ~= a^3 / M
~= (7 x 10^2) / 20
~= 3.5 x 10^1
P ~= sqrt(35)
~= 6 yearsA more careful calculation gives about 5.8 years, very close to the 5.5-to-6-year period range cited in reporting on earlier orbital models.
For speed, scale from Earth's 30 km/s orbital speed:
v ~= Earth speed x sqrt(M / a)
~= 30 km/s x sqrt(20 / 8.8)
~= 30 km/s x sqrt(2.3)
~= 30 km/s x 1.5
~= 45 km/sYou can get the same answer from circumference divided by period:
speed ~= Earth speed x (orbit radius ratio / period ratio)
~= 30 km/s x (8.8 / 5.8)
~= 45 km/sIn a century, the companion could complete roughly:
orbits per century ~= 100 years / 5.8 years
~= 17 complete orbitsSo Matt's estimated 4-year period and 25 orbits per century were in the right order of magnitude. The larger miss was speed: using the too-long 4-year period and the 9-AU path should have produced a speed higher than Earth's, not 7 km/s. This is a useful check: a massive 20-solar-mass system at Saturn-like distance can still move faster than Earth because the central mass is so much larger.
The direct-detection delay is therefore not explained by the orbit being slow. Astronomers had many orbital cycles in principle. The hard part was imaging a relatively faint companion extremely close to a huge, bright, variable red supergiant, at a separation of only tens of milliarcseconds as seen from Earth. The companion moved; telescopes, contrast, timing, and image processing had to catch up.
Post-check reflection
Matt's reflection
I used the wrong model. Gravitational force was not the right equation to use. I did not know formulas for orbital acceleration and could not derive an equation that would allow me to compare the two systems.
It is funny: my answers still wound up in the right order of magnitude except when I got to the orbital speed calculation.
My only major takeaway on this one was the importance of brushing up on orbital physics.
Recommended memory peg
Remember 1 AU is the Earth-Sun distance, Earth orbits at about 30 km/s, and in AU/solar-mass/year units P^2 = a^3 / M. For orbital speed, use v ~= 30 km/s x sqrt(M/a).
Reader results
Bars show how submitted estimates sort into the answer choices from the gut-check prompt.